Build this simple circuit yourself using veroboard or stripboard - and a handful of cheap off-the-shelf components. Connect it to the output terminals of your power supply - and it will protect both your PSU - and the device you're powering-up.

COMMENTS SUGGESTIONS	Add-On Current Limiter - Support Material	MORE POWER SUPPLY CIRCUITS

Current Limiter Schematic	The Rest of Ron's Circuits	Write To Ron	More Free-to-Use Circuits	Circuit Exchange International

Circuit Description

Click Here For A Photograph Of The Prototype.

Circuit Diagram For An
Add-on Current Limiter

This circuit places a maximum limit on the current available from the output terminals.

It does this by limiting the maximum current that can flow through Q1.

Because the output terminals are in series with Q1 - limiting the Q1 current also limits the output current.

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The negative output terminal is connected to the collector of Q1.

So any current drawn from the positive terminal - flows into the collector of Q1.

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Thus - the Q1 collector current is supplied by the output terminals.

The Q1 base current is supplied by R1. 

And the Q1 emitter current - flowing to ground through R2 - is the sum of the collector current and the base current.

Because the base current is relatively small - for present purposes - it can be ignored.

And you can think of the emitter current as being equal to the collector current.

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Or - to put it the other way round - the collector current is equal to the emitter current.

So - if you set a limit on the size of the emitter current - you also set a limit on the size of the collector current.

That is - you set a limit on the amount of current available from the output terminals.

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The key to limiting the Q1 emitter current - is to set a maximum limit on the transistor's base voltage. 

That's the purpose of D1 & D2

Each diode has a forward voltage drop of about 0v7.

The two together prevent the base voltage of Q1 from rising above about 1v4.

It may be lower than 1v4 - but it cannot go higher than 1v4.

In effect - D1 & D2 are acting like a 1v4 zener diode.

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The voltage on the base of a silicon NPN transistor is always about 0v7 higher than the voltage on its emitter.

So - when D1 & D2 prevent the voltage on the base of Q1 from rising above 1v4 - they also prevent the voltage across R2 from rising above about 0v7.

It may be lower than 0v7 - but it cannot go higher than 0v7.

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If the maximum possible voltage across R2 is 0v7 - then the maximum possible current through it is 

                     0.7 ÷ R2.

In the drawing - R2 is a 10 ohm resistor.

And the maximum current that will flow through it is: - 

                 0.7 ÷ 10 = 70mA.

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R2 is connected in series with Q1 and the output terminals.

If 70mA is the largest current that can flow through R2 - it's also the largest current that will flow through the transistor.

And - it's the maximum current that will flow between the output terminals - even if they're shorted together.

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To understand how this works - first consider what would happen if the diodes were not present - and the output terminals were shorted together.

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As the base current through R1 begins to turn the transistor on - current will start to flow through both the transistor and R2. 

So the voltage across R2 will begin to rise. 

This causes the voltage on the emitter to rise. 

And - since the base is always 0v7 higher than the emitter - the voltage on the base is carried upwards by the emitter.

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Eventually, virtually the whole of the supply voltage will appears across R2. 

So - with a 12-volt supply - the maximum current flowing through the output terminals, the transistor and R2 is roughly

      (V ÷ R) = (12 ÷ 10) = 1.2 amps

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Now consider the same situation with D1 & D2 in place. 

The two diodes prevent the base of Q1 from rising above 1v4. 

This means that the base voltage cannot ride up on the back of the emitter.

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The base cannot go higher than 1v4. 

So the emitter voltage cannot go higher than 0v7. 

And the highest current that will flow between the output terminals - is one that cause a drop of 0v7 across R2. 

      That is:  I(max) = (V ÷ R2) 

         = (0.7 ÷ 10) = 70mA.

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Consider what would happen if the voltage on the emitter tended to rise above 0v7. 

The base-emitter junction would no longer be forward biased. 

So the transistor would begin to turn off. 

This would lead to a fall in the current flowing through R2 - resulting in a drop in the voltage across it. 

This drop in voltage would restore the forward bias to the base-emitter junction - and the current through R2 would rise again.

In other words - the voltage across R2 is controlling the transistor. 

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While the current demanded is below the limit - the voltage on the emitter of Q1 is below 0v7 - and the transistor is turned fully on. 

Except for the small loss across R2 - the output voltage is virtually the same as the input voltage.

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If the load on the output terminals becomes excessive - and the voltage on the emitter of Q1 reaches 0v7 - the transistor will begin to turn off. 

As it does so - the voltage at the collector of Q1 will rise. This reduces the output voltage. 

The reduced output voltage - reduces the load on the output terminals.

And the output current is maintained at the maximum level fixed by R2. 

That is - at the level required to produce a voltage of 0v7 at the emitter of Q1. 

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Because the output voltage falls - a voltmeter connected across the output terminals - will tell you if the load is attempting to draw too much current.

If you don't want to use a voltmeter - you can use a 2k7 resistor and an LED instead. 

The standby current flowing through the LED is already relatively small.

Any fall in the output voltage - will cause the LED current to fall even further. 

So the LED will dim - or even extinguish completely.

This should give an adequate visual indication - when the load on the output is excessive.

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There's no advantage in using extra diodes to raise the Q1 base voltage above 1v4.

The only effect it has is to raise the voltage drop across R2.

This means that R2 has to dissipate more energy - i.e. watts.

And the gap between the input and output voltages is increased.

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       How to change the limit.

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When you build an electronic project - it's easy to make a mistake.

When you power up the circuit for the first time - the result is often a very expensive puff of smoke.

I wanted a cheap add-on device that would limit the maximum current available.

And - hopefully - limit the damage caused.

To this end - I considered limits below 100mA or so to be adequate.

I haven't built the circuit with limits above 100mA.

But - if you wish to do so - the following may help.

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To work out the value of the resistor required to produce a particular limit - you need an equation for R2.
 
         If  Current = (0.7 ÷ R2)
 
        Multiply both sides by R2
 
           (R2 x Current) = 0.7
 
      Divide both sides by Current
 
            R2 = (0.7 ÷ Current)
 
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For a limit of 200mA - R2 needs to be

         (0.7 ÷ 0.2) = 3.5 ohms
 
For a limit of 500mA - R2 needs to be

          (0.7 ÷ 0.5) = 1.4 ohms

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If more or less will do - just use the nearest preferred value. i.e. 3R3 - 1R5

If you require more accuracy - connect a number of resistors in parallel.

For example - connecting a 22 ohm resistor in parallel with the 1R5 will reduce the resistance to 1R4.

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It's better to construct your parallel resistor network by trial and error - rather than to rely on calculations.

There are a number of reasons for this.

Firstly, the actual voltage across the individual diodes - and the base-emitter junction of Q1 - will not be exactly 0v7.

Secondly, some of the output current is consumed by R3 and the LED .

And finally, the base current - although relatively small - does contribute to the voltage drop across R2.

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If you increase the current - you'll also increase the watts.

        Watts = Current x Volts

The voltage across R2 is 0v7

So - with a current of 70mA - R2 has to dissipate 
        
        0.07 x 0.7 = about 50mW.

Increase the current to 500mA and R2 has to dissipate 

        0.5 x 0.7 = about 350mW.

I specified a 1-watt resistor for R2 - because it will cover a wide range of limits.

It's a sort of One-Size-Fits-All solution.

Depending on the actual limit - a lower wattage resistor may be adequate.

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As the current limit is increased - the gain of Q1 may become important.

For Example: -

At 500mA the minimum DC current gain of a BD131 is listed as 40.

This is the guaranteed minimum gain. The actual gain is likely to be higher.

However - you could need a base current of up to 

           500 ÷ 40 = 12.5mA.

So - particularly with a lower input voltage - you may need to reduce the value of R1.

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As the current limit is increased - the amount of power Q1 has to dissipate may become important.

For Example: -

The worst possible case is that the output terminals are shorted together.

For practical purposes - this means that the full supply voltage is across Q1.

If the supply is 12-volt and the limit is 500mA - Q1 will have to dissipate 

          12 x 0.5 = 6-watts.

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It's important to understand the relationship between a transistor's power rating - and its operating temperature.

As a transistor's operating temperature rises - the maximum power it can handle falls. 

From the Data Sheet - the BD131's absolute maximum power rating is given as 15 watts.

However - this rating only applies if the operating temperature is below 60°C.

In other words - the BD131 will handle up to 15 watts.

But only if it's bolted to a heatsink - big enough to keep the operating temperature below 60°C.

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       How the alarm circuit works.

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If the load on the output terminals is excessive - the output voltage falls.

That is - the potential difference across the output terminals falls.

The potential difference falls - because the Q1 collector voltage rises.

Thus - an increase in the collector voltage is an indication that the load on the output is excessive. 

The alarm circuit uses this increase in collector voltage to sound the Buzzer.

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In standby mode Q2 is switched firmly off by R5 - and the Buzzer is silent. 

The diodes - D3 & D4 - are wired in series with R4 and the base-emitter junction of Q2. 

D3, D4 and the base-emitter junction - each have a forward voltage drop of about 0v7.

So the three together represent a forward voltage threshold of about 
2-volts.

That is - roughly 0v7 x 3.

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This threshold must be overcome before base current will flow into Q2 - turn the transistor on - and sound the Buzzer. 

In other words - the alarm will only sound if the Q1 collector voltage rises above about 2-volts.

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When the Q1 collector voltage exceeds 2 volts - base current will flow into Q2.

This will switch the transistor on.

Q2 will connect the negative lead of the Buzzer to ground.

And the Buzzer will sound. 

If you want the alarm to kick in earlier - just use one diode in series with R4. 

This reduces the threshold to about 1v4.

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Parts List

Suppliers

Worldwide RS Components

UK & Ireland Maplin

Construction Notes

Click here if you're new to constructing stripboard projects

The terminals are a good set of reference points. To fit them, you may need to enlarge the holes slightly. Then turn the board over and use a felt-tip pen to mark the 4 places where the tracks are to be cut. Before you cut the tracks - double check that your marks are in the right place.

When you're satisfied that the pattern is right - cut the tracks. Make sure that the copper is cut all the way through. Sometimes a small strand of copper remains at the side of the cut and this will cause malfunction. Use a magnifying glass - and backlight the board. It only takes the smallest strand of copper to cause a problem. If you don't have the proper track-cutting tool, then a 6 to 8mm drill-bit will do. Just use the drill-bit as a hand tool - there's no need for a drilling machine.

Make 1

Start by making and fitting the Four Wire Links. I used bare copper wire on the component side of the board. Telephone cable is suitable - the single stranded variety used indoors to wire telephone sockets. Stretching the core slightly will straighten it - and also allow the insulation to slip off.

Next - fit the 5 resistors. The resistors are all shown lying flat on the board. But those connected between close or adjacent tracks are actually mounted standing upright.

Then fit the five diodes - the BC547 - and the LED. Pay particular attention to the orientation of the diodes. Note that D5 - across the Buzzer terminals - faces in the opposite direction to the other diodes. Again, the diodes are all shown lying flat on the board - but those connected between adjacent tracks are actually mounted standing upright.

Make 2

Finally, fit the BD131 and the heatsink. If you have set a high maximum current limit - and the output may be overloaded for long periods of time - you should increase the size of the heatsink. The transistor has to dissipate most energy when the output terminals are shorted together. Try it. If the BD131 is getting too hot to touch - your heatsink is inadequate.

Current Limiter Schematic	The Rest of Ron's Circuits	Write To Ron	More Free-to-Use Circuits	Circuit Exchange International

Add-On Current Limiter - Support Material

Construction Notes